[tex]f(x)=2x^2-8x=2x(x-4)\\\\x-intercept\to f(x)=0\\\\2x(x-4)=0\iff2x=0\ \vee\ x-4=0\\\\\boxed{x_1=0\ \vee\ x_2=4}[/tex]
[tex]\text{The vertex of}\ f(x)=ax^2+bx+c:\\\\(h,\ k)\to h=\dfrac{-b}{2a},\ k=f(h)\\\\\text{We have}\ f(x)=2x^2-8x\\\\a=2,\ b=-8,\ c=0\\\\\text{substitute}\\\\h=\dfrac{-(-8)}{2\cdot2}=\dfrac{8}{4}=2\\\\k=f(2)=2(2^2)-8(2)=2(4)-16=8-16=-8\\\\\boxed{The\ vertex\ (2,\ -8)}[/tex]
